# Make three non-empty sets with negative, positive and 0 products

• Difficulty Level : Medium
• Last Updated : 22 Jun, 2021

You are given an array of n distinct integers. Your task is to divide this array into three non-empty sets so as the following conditions hold:

1. The first set should contain the elements such that their product should be less than 0.
2. The Second set should contain the elements such that their product should be greater than 0.
3. The Third set should contain the elements such that their product should be equal to 0.

Notes:-

1. In the give array, each number must occur in exactly one set.
2. It may be assumed that we can always make three sets (there is at-least one negative element and one 0 in input array).

Examples:

```Input : 4
arr[] = -1 -2 -3 0
Output :  -1
-3 -2
0
In this example, product of first
set is negative, product of second
set is positive and product of third
set is 0.```

In this problem we just need to Split input data into 3 vectors: first will contain negative numbers, second positive numbers, third zeroes. If size of first vector is even move one number from it to the third vector. If second vector contains only 1, then move two numbers from first vector to the second vector.

## C++

 `// CPP program to make three non-empty sets``// as per the given conditions.``#include ``using` `namespace` `std;` `void` `makeSets(``int` `arr[], ``int` `n)``{``    ``vector<``int``> first, second, third;` `    ``// insert number equal to 0 to third set.``    ``// numbers greater than 0 to second set.``    ``// insert numbers less than 0 to first set.``    ``for` `(``int` `i = 0; i < n; i++) {``        ``if` `(arr[i] == 0)``            ``third.push_back(arr[i]);``        ``if` `(arr[i] > 0)``            ``second.push_back(arr[i]);``        ``if` `(arr[i] < 0)``            ``first.push_back(arr[i]);``    ``}` `    ``if` `(first.size() == 0 || third.size() == 0)``    ``{``        ``cout << ``"Not Possible"``;``        ``return``;``    ``}` `    ``// if second set is empty.``    ``if` `(second.size() == 0) {``        ``for` `(``int` `i = 0; i < 2; i++)``        ``{``            ``second.push_back(first.back());``            ``first.pop_back();``        ``}``    ``}` `    ``// if length of first set is even.``    ``if` `(first.size() % 2 == 0) {``        ``third.push_back(first.back());``        ``first.pop_back();``    ``}` `    ``// output the first set elements.``    ``for` `(``int` `i = 0; i < first.size(); i++)``        ``cout << first[i] << ``" "``;   ``    ` `    ``// output the second set elements.``    ``cout << endl;``    ``for` `(``int` `i = 0; i < second.size(); i++)``        ``cout << second[i] << ``" "``;   ` `    ``// output the third set elements.``    ``cout << endl;``    ``for` `(``int` `i = 0; i < third.size(); i++)``        ``cout << third[i] << ``" "``;   ``}` `// Driver Function``int` `main()``{``    ``int` `arr[] = { -1, 2, 0 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);``    ``makeSets(arr, n);``    ``return` `0;``}`

## Java

 `// Java program to make three non-empty sets``// as per the given condition.``import` `java.util.*;` `class` `GFG``{` `    ``static` `void` `makeSets(``int` `arr[], ``int` `n)``    ``{``        ``Vector first = ``new` `Vector();``        ``Vector second = ``new` `Vector();``        ``Vector third = ``new` `Vector();` `        ``// insert number equal to 0 to third set.``        ``// numbers greater than 0 to second set.``        ``// insert numbers less than 0 to first set.``        ``for` `(``int` `i = ``0``; i < n; i++)``        ``{``            ``if` `(arr[i] == ``0``)``            ``{``                ``third.add(arr[i]);``            ``}``            ``if` `(arr[i] > ``0``)``            ``{``                ``second.add(arr[i]);``            ``}``            ``if` `(arr[i] < ``0``)``            ``{``                ``first.add(arr[i]);``            ``}``        ``}` `        ``if` `(first.size() == ``0` `|| third.size() == ``0``)``        ``{``            ``System.out.print(``"Not Possible"``);``            ``return``;``        ``}` `        ``// if second set is empty.``        ``if` `(second.size() == ``0``)``        ``{``            ``for` `(``int` `i = ``0``; i < ``2``; i++)``            ``{``                ``second.add(first.lastElement());``                ``first.remove(first.size() - ``1``);``            ``}``        ``}` `        ``// if length of first set is even.``        ``if` `(first.size() % ``2` `== ``0``)``        ``{``            ``third.add(first.lastElement());``            ``first.remove(first.size() - ``1``);``        ``}` `        ``// output the first set elements.``        ``for` `(``int` `i = ``0``; i < first.size(); i++)``        ``{``            ``System.out.print(first.get(i) + ``" "``);``        ``}` `        ``// output the second set elements.``        ``System.out.println();``        ``for` `(``int` `i = ``0``; i < second.size(); i++)``        ``{``            ``System.out.print(second.get(i) + ``" "``);``        ``}` `        ``// output the third set elements.``        ``System.out.println();``        ``for` `(``int` `i = ``0``; i < third.size(); i++)``        ``{``            ``System.out.print(third.get(i) + ``" "``);``        ``}``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[] = {-``1``, ``2``, ``0``};``        ``int` `n = arr.length;``        ``makeSets(arr, n);``    ``}``}` `/* This code contributed by PrinciRaj1992 */`

## Python3

 `# Python3 program to make three non-empty sets``# as per the given conditions.``def` `makeSets(arr, n):``    ``first ``=` `[]``    ``second ``=` `[]``    ``third ``=` `[]``    ` `    ``# insert number equal to 0 to third set.``    ``# numbers greater than 0 to second set.``    ``# insert numbers less than 0 to first set.``    ``for` `i ``in` `range``(n):``        ``if` `(arr[i] ``=``=` `0``):``            ``third.append(arr[i])``        ``if` `(arr[i] > ``0``):``            ``second.append(arr[i])``        ``if` `(arr[i] < ``0``):``            ``first.append(arr[i])``    ` `    ``if` `(``len``(first) ``=``=` `0` `or` `len``(third) ``=``=` `0``):``        ``print``(``"Not Possible"``)``        ``return``    ` `    ``# if second set is empty.``    ``if` `(``len``(second)``=``=` `0``):``        ``for` `i ``in` `range``(``2``):``            ``second.append(first[``-``1``])``            ``first.pop()``            ` `    ``# if length of first set is even.``    ``if` `(``len``(first) ``%` `2` `=``=` `0``):``        ``third.append(first[``-``1``])``        ``first.pop()``    ` `    ``# output the first set elements.``    ``for` `i ``in` `range``(``len``(first)):``        ``print``(first[i], end ``=` `" "``)``    ` `    ``# output the second set elements.``    ``print``()``    ``for` `i ``in` `range``(``len``(second)):``        ``print``(second[i], end ``=` `" "``)``        ` `    ``# output the third set elements.``    ``print``()``    ``for` `i ``in` `range``(``len``(third)):``        ``print``(third[i], end ``=` `" "``)``        ` `# Driver Function``arr ``=` `[``-``1``, ``2``, ``0``]``n ``=` `len``(arr)``makeSets(arr, n)` `# This Code is contributed by SHUBHAMSINGH10`

## C#

 `// C# program to make three non-empty sets``// as per the given condition.``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{` `    ``static` `void` `makeSets(``int` `[]arr, ``int` `n)``    ``{``        ``List<``int``> first = ``new` `List<``int``>();``        ``List<``int``> second = ``new` `List<``int``>();``        ``List<``int``> third = ``new` `List<``int``>();` `        ``// insert number equal to 0 to third set.``        ``// numbers greater than 0 to second set.``        ``// insert numbers less than 0 to first set.``        ``for` `(``int` `i = 0; i < n; i++)``        ``{``            ``if` `(arr[i] == 0)``            ``{``                ``third.Add(arr[i]);``            ``}``            ``if` `(arr[i] > 0)``            ``{``                ``second.Add(arr[i]);``            ``}``            ``if` `(arr[i] < 0)``            ``{``                ``first.Add(arr[i]);``            ``}``        ``}` `        ``if` `(first.Count == 0 || third.Count == 0)``        ``{``            ``Console.Write(``"Not Possible"``);``            ``return``;``        ``}` `        ``// if second set is empty.``        ``if` `(second.Count == 0)``        ``{``            ``for` `(``int` `i = 0; i < 2; i++)``            ``{``                ``second.Add(first[first.Count-1]);``                ``first.Remove(first.Count - 1);``            ``}``        ``}` `        ``// if length of first set is even.``        ``if` `(first.Count % 2 == 0)``        ``{``            ``third.Add(first[first.Count-1]);``            ``first.Remove(first.Count - 1);``        ``}` `        ``// output the first set elements.``        ``for` `(``int` `i = 0; i < first.Count; i++)``        ``{``            ``Console.Write(first[i] + ``" "``);``        ``}` `        ``// output the second set elements.``        ``Console.WriteLine();``        ``for` `(``int` `i = 0; i < second.Count; i++)``        ``{``            ``Console.Write(second[i] + ``" "``);``        ``}` `        ``// output the third set elements.``    ``Console.WriteLine();``        ``for` `(``int` `i = 0; i < third.Count; i++)``        ``{``            ``Console.Write(third[i] + ``" "``);``        ``}``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``int` `[]arr = {-1, 2, 0};``        ``int` `n = arr.Length;``        ``makeSets(arr, n);``    ``}``}` `// This code has been contributed by 29AjayKumar`

## Javascript

 ``

Output:

```-1
2
0```

Time Complexity :- O(n)

My Personal Notes arrow_drop_up